Super User is a question and answer site for computer enthusiasts and power users. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

According to this article , the bus is the circuit that connects the motherboard . The faster the speed of the bus , the quicker data can be transferred . Bus speed is measured in MHz (Mega Hertz) and a bus speed of 66 MHZ means that that are 66 million cycles per second . Hertz means how the frequency of x every second

For CPU , 66 MHz means 66 million instructions can be processed every second

How about for Bus ?? How much data is terms of btyes is transferred for every cycle ??

share|improve this question
"For CPU , 66 MHz means 66 million instructions can be processed every second" -- That's an incorrect assumption, more than one instruction can be performed per clock cycle. As for your question, the size of the bus matters. IE: a 32-bit data bus can move less data at once than a 64-bit data bus could. – Ƭᴇcʜιᴇ007 Mar 12 '14 at 17:48
When it comes to frequency considerations, there are other factors to be considered... Such as bus widths. Simply, a pin can change state 66 million times in a second at (66 MHz). – Ghassan Mar 12 '14 at 17:51
For the amount of data (in bytes) being sent, it depends on how big the bus is (in bits). If you have a 32-bit bus, you can send 4 Bytes of information per cycle (8 bits / 1 byte). – user304064 Mar 12 '14 at 17:52
To continue on what @techie007 said, it is not limited to more than one CPU instuction per cycle. DDR Memory will do two data transfers per one clock cycle. – Scott Chamberlain Mar 12 '14 at 18:22

It's the speed at which the PCI protocol operates over the bus. It's the CLK signal in the illustration below.

Wikipedia gives this formula for 32-bit 33MHz PCI, which is x2 for 32-bit 64MHz PCI:

Peak transfer rate of 133 MB/s (133 megabytes per second) for 32-bit bus width (33.33 MHz × 32 bits ÷ 8 bits/byte = 133 MB/s)

There is an address phase and a data phase in the protocol as you can see below, so not all of the 266MBytes/sec will be dedicated to actual work. There is a bit of overhead.

From this:

The basic transfer mechanism is a burst, composed of an address phase and one or more data phases.

Typical read and write transfers are illustrated below:

enter image description here

This article mentions an effective throughput of 90MBytes/sec of the theoretical 133MBytes/sec due to the overhead. For 32-bit 66MHz it would be doubled.

share|improve this answer

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .